Scala在资源文件夹中获取文件的文件路径 [英] Scala get file path of file in resources folder

问题描述

我使用的是Stanford CRFClassifier，为了运行，它需要一个训练有素的分类器模型。我把这个文件放在资源目录中。来自CRFClassifier的Javadocs http://nlp.stanford.edu/nlp/javadoc/javanlp/edu/stanford/nlp/ie/crf/CRFClassifier.html#getClassifier(java.lang.String)
文件的路径必须是CRFClassifier.getClassifier（）的输入，它是一个java.lang.String对象。所以我的问题是如何告诉.getClassifier（）文件在资源目录中？即如何获取资源目录中文件的文件路径？

I am using the Stanford CRFClassifier and in order to run, it requires a file that is the trained classifier model. I have put this file in the resources directory. From the Javadocs for the CRFClassifier http://nlp.stanford.edu/nlp/javadoc/javanlp/edu/stanford/nlp/ie/crf/CRFClassifier.html#getClassifier(java.lang.String) the path to the file must be an input to CRFClassifier.getClassifier() and it is a java.lang.String object. So my question is how do I tell .getClassifier() that the file is in the resources directory? i.e. how do I get the file path of a file in the resources directory?

我试过简单地

val classifier = CRFClassifier.getClassifier("./src/main/resources/my_model.ser.gz")

但是这会返回一个FileNotFoundException。

But this returns a FileNotFoundException.

我也试过了

Source.fromURL(getClass.getResource("/my_model.ser.gz"))

返回一个BufferedSource对象，但我不知道如何从中获取文件路径。

which returns a BufferedSource object, but I do not know how to get a file path from this.

任何帮助都将不胜感激。

Any help would be greatly appreciated.

推荐答案

我设法通过以下方式获取文件路径

I managed to be able to get the file path by doing the following

val url = getClass.getResource（/ my_model.ser.gz）

val classifier = CRFClassifier.getClassifier（url.getPath（））

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